3.593 \(\int \sqrt {a+(c \cos (e+f x)+b \sin (e+f x))^2} \, dx\)

Optimal. Leaf size=79 \[ \frac {\sqrt {a+(b \sin (e+f x)+c \cos (e+f x))^2} E\left (e+f x+\tan ^{-1}(b,c)|-\frac {b^2+c^2}{a}\right )}{f \sqrt {\frac {(b \sin (e+f x)+c \cos (e+f x))^2}{a}+1}} \]

[Out]

(cos(e+f*x+arctan(b,c))^2)^(1/2)/cos(e+f*x+arctan(b,c))*EllipticE(sin(e+f*x+arctan(b,c)),((-b^2-c^2)/a)^(1/2))
*(a+(c*cos(f*x+e)+b*sin(f*x+e))^2)^(1/2)/f/(1+(c*cos(f*x+e)+b*sin(f*x+e))^2/a)^(1/2)

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Rubi [F]  time = 0.70, antiderivative size = 0, normalized size of antiderivative = 0.00, number of steps used = 0, number of rules used = 0, integrand size = 0, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.000, Rules used = {} \[ \int \sqrt {a+(c \cos (e+f x)+b \sin (e+f x))^2} \, dx \]

Verification is Not applicable to the result.

[In]

Int[Sqrt[a + (c*Cos[e + f*x] + b*Sin[e + f*x])^2],x]

[Out]

((I/2)*Defer[Subst][Defer[Int][Sqrt[a + (c + b*x)^2/(1 + x^2)]/(I - x), x], x, Tan[e + f*x]])/f + ((I/2)*Defer
[Subst][Defer[Int][Sqrt[a + (c + b*x)^2/(1 + x^2)]/(I + x), x], x, Tan[e + f*x]])/f

Rubi steps

\begin {align*} \int \sqrt {a+(c \cos (e+f x)+b \sin (e+f x))^2} \, dx &=\frac {\operatorname {Subst}\left (\int \frac {\sqrt {a+\frac {(c+b x)^2}{1+x^2}}}{1+x^2} \, dx,x,\tan (e+f x)\right )}{f}\\ &=\frac {\operatorname {Subst}\left (\int \left (\frac {i \sqrt {a+\frac {(c+b x)^2}{1+x^2}}}{2 (i-x)}+\frac {i \sqrt {a+\frac {(c+b x)^2}{1+x^2}}}{2 (i+x)}\right ) \, dx,x,\tan (e+f x)\right )}{f}\\ &=\frac {i \operatorname {Subst}\left (\int \frac {\sqrt {a+\frac {(c+b x)^2}{1+x^2}}}{i-x} \, dx,x,\tan (e+f x)\right )}{2 f}+\frac {i \operatorname {Subst}\left (\int \frac {\sqrt {a+\frac {(c+b x)^2}{1+x^2}}}{i+x} \, dx,x,\tan (e+f x)\right )}{2 f}\\ \end {align*}

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Mathematica [B]  time = 1.67, size = 325, normalized size = 4.11 \[ -\frac {\left (\left (b^2-c^2\right ) \sin (2 (e+f x))+2 b c \cos (2 (e+f x))\right ) \sqrt {2 a+\left (c^2-b^2\right ) \cos (2 (e+f x))+b^2+2 b c \sin (2 (e+f x))+c^2} E\left (\sin ^{-1}\left (\frac {\sqrt {\frac {\left (b^2-c^2\right ) \cos (2 (e+f x))-2 b c \sin (2 (e+f x))+\sqrt {\left (b^2+c^2\right )^2}}{\sqrt {\left (b^2+c^2\right )^2}}}}{\sqrt {2}}\right )|\frac {2 \sqrt {\left (b^2+c^2\right )^2}}{b^2+c^2+2 a+\sqrt {\left (b^2+c^2\right )^2}}\right )}{\sqrt {2} f \sqrt {\left (b^2+c^2\right )^2} \sqrt {\frac {\left (\left (b^2-c^2\right ) \sin (2 (e+f x))+2 b c \cos (2 (e+f x))\right )^2}{\left (b^2+c^2\right )^2}} \sqrt {\frac {2 a+\left (c^2-b^2\right ) \cos (2 (e+f x))+b^2+2 b c \sin (2 (e+f x))+c^2}{2 a+\sqrt {\left (b^2+c^2\right )^2}+b^2+c^2}}} \]

Antiderivative was successfully verified.

[In]

Integrate[Sqrt[a + (c*Cos[e + f*x] + b*Sin[e + f*x])^2],x]

[Out]

-((EllipticE[ArcSin[Sqrt[(Sqrt[(b^2 + c^2)^2] + (b^2 - c^2)*Cos[2*(e + f*x)] - 2*b*c*Sin[2*(e + f*x)])/Sqrt[(b
^2 + c^2)^2]]/Sqrt[2]], (2*Sqrt[(b^2 + c^2)^2])/(2*a + b^2 + c^2 + Sqrt[(b^2 + c^2)^2])]*Sqrt[2*a + b^2 + c^2
+ (-b^2 + c^2)*Cos[2*(e + f*x)] + 2*b*c*Sin[2*(e + f*x)]]*(2*b*c*Cos[2*(e + f*x)] + (b^2 - c^2)*Sin[2*(e + f*x
)]))/(Sqrt[2]*Sqrt[(b^2 + c^2)^2]*f*Sqrt[(2*a + b^2 + c^2 + (-b^2 + c^2)*Cos[2*(e + f*x)] + 2*b*c*Sin[2*(e + f
*x)])/(2*a + b^2 + c^2 + Sqrt[(b^2 + c^2)^2])]*Sqrt[(2*b*c*Cos[2*(e + f*x)] + (b^2 - c^2)*Sin[2*(e + f*x)])^2/
(b^2 + c^2)^2]))

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fricas [F]  time = 0.51, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\sqrt {2 \, b c \cos \left (f x + e\right ) \sin \left (f x + e\right ) - {\left (b^{2} - c^{2}\right )} \cos \left (f x + e\right )^{2} + b^{2} + a}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+(c*cos(f*x+e)+b*sin(f*x+e))^2)^(1/2),x, algorithm="fricas")

[Out]

integral(sqrt(2*b*c*cos(f*x + e)*sin(f*x + e) - (b^2 - c^2)*cos(f*x + e)^2 + b^2 + a), x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \sqrt {{\left (c \cos \left (f x + e\right ) + b \sin \left (f x + e\right )\right )}^{2} + a}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+(c*cos(f*x+e)+b*sin(f*x+e))^2)^(1/2),x, algorithm="giac")

[Out]

integrate(sqrt((c*cos(f*x + e) + b*sin(f*x + e))^2 + a), x)

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maple [B]  time = 4.68, size = 4067972, normalized size = 51493.32 \[ \text {output too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+(c*cos(f*x+e)+b*sin(f*x+e))^2)^(1/2),x)

[Out]

result too large to display

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \sqrt {{\left (c \cos \left (f x + e\right ) + b \sin \left (f x + e\right )\right )}^{2} + a}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+(c*cos(f*x+e)+b*sin(f*x+e))^2)^(1/2),x, algorithm="maxima")

[Out]

integrate(sqrt((c*cos(f*x + e) + b*sin(f*x + e))^2 + a), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \sqrt {a+{\left (c\,\cos \left (e+f\,x\right )+b\,\sin \left (e+f\,x\right )\right )}^2} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + (c*cos(e + f*x) + b*sin(e + f*x))^2)^(1/2),x)

[Out]

int((a + (c*cos(e + f*x) + b*sin(e + f*x))^2)^(1/2), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \sqrt {a + \left (b \sin {\left (e + f x \right )} + c \cos {\left (e + f x \right )}\right )^{2}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+(c*cos(f*x+e)+b*sin(f*x+e))**2)**(1/2),x)

[Out]

Integral(sqrt(a + (b*sin(e + f*x) + c*cos(e + f*x))**2), x)

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